Is $f: [0,1[ \cup \{ 2 \} \to [0,1]$ continuous?

Question

I am having this transformation $f: [0,1[ \cup \{ 2 \} \to [0,1]$

$$f(x) = \begin{cases} x & x \neq 2 \\1 & x = 2 \end{cases}$$

How can I prove that this transformation is continuous or not continuous with the epsilon-delta definition?

Answer 1

First, the function is continuous on $[0,1[$.

The only question is if the function is continuous at $2$. The usual definition is that for all $\epsilon > 0$ you have to be able to find a $\delta > 0$ such that if $0< \lvert x- 2 \rvert < \delta$ then $\lvert f(x) - 1\rvert < \epsilon$. For a given $\epsilon$ you can simply pick $\delta = \frac{1}{2}$. Then for $0<\lvert x - 2 \rvert < \frac{1}{2}$ you will have $\lvert f(x) - 1\rvert < \epsilon$.

So the function is continuous at $2$.

Answer 2

Your function $f$ is a restriction of $g(x)=3/2-|x-3/2|$ and so is continuous.